I want to know how to determine det(B) if det(A) = 3? 1 a 1 4 2a+1 2+b A = 2 1 b B = 15 5c 0...
I want to know how to determine det(B) if det(A) = 3?
1 
a 
1 
4 
2a+1 
2+b 

A = 
2 
1 
b 
B = 
15 
5c 
0 

3 
C 
0 
2 
1 
b 
2 Answers  Add Yours
First we will write the formula for det (A)
det(A) = 1*1*0 + a*b*3 + 2*c* *1  1*1*3  a*2*0 c *b*1 = 3
Now simplify:
det(A) = 3ab + 2c  3 bc = 3
= 3ab + 2c  bc = 6............(1)
Now we will calculate det(B):
det(B) = 4*5c*b + (2a+1)*0*2 + (2+b)*15*1  (2+b)*5c*2  (2a+1)*15*b  4*0*1
= 20bc + 30 + 15b  20c  10bc  30ab  15b
= 10bc  20c + 30 30ab
Let us rewrite:
==>det(B) = 30ab  20c + 10bc + 30
Factor = 10:
==> det (B) = 10 (3ab + 2c  bc) + 30
But from (1) we know that (3ab+2c  bc) = 6
==> det(B) = 10*6 + 30
= 60 + 30 = 30
==> det(B) = 30
Expandinfg thruogh the 1st row, we get :
Det (A) = [(1 a 1),(2 1 b), (3 c 0)] = 3......(1)
Det B = [(4 2a+1 2+b ) , (15 5c 0) , (2 1 b)].
Det B = R1R3 = [(42 2a+11 2+bb) , (15 5c 0), (2 1 b)].
DetB = [(2 2a 2), (15 5c 0) , (2 1 b)]
Det B = 2 *5 [(1 a 1), (2 c 0) , (2 1 b)] , where 2 and 5 are factored out from 1st and 2nd rows.
Now we inter change R2 and R3 which results
Det B = 10 [( 1 a 1), (2 1 b) , ( 2 c 0)].....(2)
Now we notice from the right side of eq (1) and left eq (2) that:
Det (B) = 10 det (A). But det(A) = 3.
Therefore det(B) = 10*3 = 30.
Therefore det(B) = 30